Tangents are drawn at three points P(t_1), Q( | vedclass

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(C) For a parabola $y^2 = 4ax$,the tangent at point $t$ is given by $ty = x + at^2$. Here,$4a = 1$,so $a = 1/4$.The intersection point of tangents at

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$7.5$

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(C) For a parabola $y^2 = 4ax$,the tangent at point $t$ is given by $ty = x + at^2$. Here,$4a = 1$,so $a = 1/4$.The intersection point of tangents at $t_i$ and $t_j$ is $(at_it_j, a(t_i + t_j))$.Given $t_1 = 2, t_2 = -4, t_3 = 6$ and $a = 1/4$:Point $L$ (intersection of $t_1, t_2$) = $(1/4(2)(-4), 1/4(2-4)) = (-2, -0.5)$.Point $M$ (intersection of $t_2, t_3$) = $(1/4(-4)(6), 1/4(-4+6)) = (-6, 0.5)$.Point $N$ (intersection of $t_3, t_1$) = $(1/4(6)(2), 1/4(6+2)) = (3, 2)$.The area of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area = $\frac{1}{2} |(-2)(0.5 - 2) + (-6)(2 - (-0.5)) + 3(-0.5 - 0.5)|$.Area = $\frac{1}{2} |(-2)(-1.5) + (-6)(2.5) + 3(-1)|$.Area = $\frac{1}{2} |3 - 15 - 3| = \frac{1}{2} |-15| = 7.5$.

Tangents are drawn at three points $P(t_1), Q(t_2), R(t_3)$ on the parabola $y^2 = x$. Let these tangents intersect each other at the points $L, M, N$. If $t_1 = 2, t_2 = -4, t_3 = 6$,then the area of the triangle $LMN$ is

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